Let P(x, y, z) be any point Then
\(PA=\sqrt { (x-3)^{ 2 }+(y-4)^{ 2 }+(z-5)^{ 2 } } \)
\(=\sqrt { x^{ 2 }+9-6x+y^{ 2 }+16-8y+z^{ 2 }+25-10z } \)
\(PA=\sqrt { (x+1)^{ 2 }+(y-3)^{ 2 }+(z+7)^{ 2 } } \)
\(\sqrt { x^{ 2 }+1+2x+y^{ 2 }+9-6y+z^{ 2 }+49-14z }\)
Now PA² + PB² = K²

Let Q(O,y, 0) be any point on y-axis. Then
\(PQ=\sqrt { (0-3)2+(y+2)2+(0-5)2 } \)
\(=\sqrt { 9+y2+4+4y+25 } \)
\(=\sqrt { y2+4y+38 } \)
But \(\sqrt { y2+4y+38 } =5\sqrt { 2 } \)
y² + 4y + 38 = 50 \(\Rightarrow \\ \) y² + 4y - 12 = 0 \(\Rightarrow \\ \) (y - 2) (y + 6) = 0
\(\Rightarrow \\ \) y = 2, -6
Thus coordinates of point Q are (0, 2, 0) and (0, -6,0).

Here P(2a, 2, 6), Q(-4, 3b, -10) and R(8, 14, 2c) are vertices of triangle PQR.
\(\therefore \)Coordinates of centroid of PQR is
\(\left( \frac { 2a-4+8 }{ 3 } ,\frac { 2+3b+14 }{ 3 } ,\frac { 6-10+2c }{ 3 } \right) \)
\(=\left( \frac { 2a+4 }{ 3 } ,\frac { 6-10+2c }{ 3 } ,\frac { 2c-4 }{ 3 } \right) \)
But is it given that coordinates of centroid is (0,0,0)
\(\therefore \frac { 2a+4 }{ 3 } =0\Rightarrow 2a+4=0\Rightarrow a+-2\)
\( \frac { 3b+16 }{ 3 } =0\Rightarrow 3b+16=0\)
\(\Rightarrow b=\frac { -16 }{ 3 } \)
\(\frac { 2c-4 }{ 3 } =0\Rightarrow 2c-4=0\Rightarrow c=2.\)

Let R(4, y, z) be any point which divides the join P(2, - 3, 4) and Q(8, 0, 10) in the ratio k: 1 internally.
\(\therefore \) coordinates of R is \(\left( \frac { 8k+2 }{ k+1 } ,\frac { -3 }{ k+1 } ,\frac { 10k+4 }{ k+1 } \right) \)
But x coordinates of R is 4
\(\therefore \frac { 8k+2 }{ k+1 } =4\Rightarrow 8k+2=4k+5\)
\(\Rightarrow k\frac { 1 }{ 2 } \)
\(\therefore y=\frac { -3 }{ \frac { 1 }{ 2 } +1 } =\frac { -3 }{ \frac { 3 }{ 2 } } =-2\)
\(z=\frac { \frac { 10\times 1 }{ 2 } +4 }{ \frac { 1 }{ 2 } +1 } =\frac { 9 }{ \frac { 3 }{ 2 } } =6\)
Thus coordinates ofR is (4, -2,6).

Here A(0,0, 6), B(0,4, 0) and C(6, 0, 0) are vertices of \(\triangle \)ABC

Now D is mid point of BC 
\(\therefore \) Coordinates of D is \(\left( \frac { 0+6 }{ 2 } ,\frac { 4+0 }{ 2 } ,\frac { 0+0 }{ 2 } \right) \)= (3,2,0)
\(\therefore \quad AD=\sqrt { (0-3)^{ 2 }+(0-2)^{ 2 }+(6-0)^{ 2 } } \)
=\(\sqrt { 9+4+36 } =\sqrt { 7 } \) units.
Also E is mid point of AC
\(\therefore \)  Coordinates of E is \(\left( \frac { 0+6 }{ 2 } ,\frac { 0+0 }{ 2 } ,\frac { 6+0 }{ 2 } \right) \)= (3,0,3)
\(\therefore \quad BE=\sqrt { (0-3)^{ 2 }+(4-0)^{ 2 }+(0-3)^{ 3 } } \)
\(=\sqrt { 9+16+9 } =\sqrt { 34 } \)units.
Also F is mid point of AB
\(\therefore \) Coordiates of F is \(\left( \frac { 0+0 }{ 2 } ,\frac { 0+4 }{ 2 } ,\frac { 6+0 }{ 2 } \right) \)=(0,2,3)
\(\therefore \) CF=\(\sqrt { (6-0)^{ 2 }+(0-2)^{ 2 }+(0-3)^{ 2 } } \)
\(=\sqrt { 36+4+9 } =7\)units.

Here A (x₁, y₁, z₁) B (x_2, y_2, z₂₎and C (x_3, y₃, z₃ )be three vertices of \(\triangle \)ABC, then coordinates of point D are
\(\left[ \frac { x1+x2+x3 }{ 2 } ,\frac { y1+y2+y3 }{ 2 } ,\frac { z1+z2+z3 }{ 2 } \right] \)

Let Gbe the centroid of ABC. Then Gdivides AD in the ratio 2 : 1. So the coordinates of G are
\(\left[ \frac { x1+2\left( \frac { x2+x3 }{ 2 } \right) }{ 1+2 } ,\frac { y1+2\left( \frac { y2+y3 }{ 2 } \right) }{ 1+2 } ,\frac { z1+2\left( \frac { z2+z3 }{ 2 } \right) }{ 1+2 } \right] \)
\(\Rightarrow \left( \frac { x1+x2+x3 }{ 3 } ,\frac { y1+y2+y3 }{ 3 } ,\frac { z1+z2+z3 }{ 3 } \right) \)

Let D(x, y, z) be the fourth vertex of parallelogram ABCD.
We know that diagonals of a parallelogram bisect each other. So the mid points ofAC and BD coincide.

\(\therefore \) Coordinates of mid point of AC \(\left( \frac { 3-1 }{ 2 } ,\frac { -1+1 }{ 2 } ,\frac { 2+2 }{ 2 } \right) \)=(1,0,2)
Also coordiantes of mid point of BD \(\left( \frac { x+1 }{ 2 } ,\frac { y+2 }{ 2 } ,\frac { z-4 }{ 2 } \right) \)
\(\therefore \quad \frac { x+1 }{ 2 } =1\Rightarrow x+1=2\Rightarrow x=1\)
\(\frac { y+2 }{ 2 } =0\Rightarrow y+2=0\Rightarrow y=-2\)
\(\frac { z-4 }{ 2 } =2\Rightarrow z-4=4\Rightarrow z=8\)
Thus the coordinates of point Dare (1, -2,8).

Let Q divides PR in the ratio k : l. Thus co-ordinates of Q are
\(\left[ \frac { 11k+5 }{ k+1 } ,\frac { 10k+4 }{ k+1 } ,\frac { -8k-2 }{ k+1 } \right] \)
It is given that coordinates of Q are (7, 6, - 4).
\(\therefore \frac { 11k+5 }{ k+1 } =7,\frac { 10k+4 }{ k+1 } =6,\frac { -8k-2 }{ k+1 } =-4\)
Now solving these, we get k=\(\frac { 1 }{ 2 } \) .
Thus Q divides PR in the ratio \(\frac { 1 }{ 2 } \) : 1 or 1 : 2.

Let P(x, y, z) be any point which is equidistant from A(3, 4, 0) and B(5, 2, -3).
Now PA = PB => PA² = PB²
\(\therefore \)(x - 3)² + (y - 4)² + (z - 0)²
= (x - 5)² + (y - 2)² + (z + 3)²
=> x²+ 9 - 6x + y² + 16 - 8y + Z²
= x² + 25 -10x + y² + 4 - 4y + Z²+ 9 + 6z
=> 4x - 4y - 6z - 13 = O.

Let A(2, 0, 3), B(O, 3, 2) and C(O, 0,1) be given points.
Let P(x, y, 0) be any point in XY plane such that PA = PB = PC.
Now PA = PB => PA2 = PB2
\(\therefore \) (x - 2)² + (y - 0)² + (- 3)²
= (x - 0)² + (y - 3)² + (- 2)²
=> x²+4-4x+y²+9
=x²+y²+9-6y+4
=> 4x - 6y = 0 => 2x - 3y = 0 ....(i)
Also PB = PC => PB² = PC²
(x - 0)² + (y - 3)² + (0 - 2)² = (x - 0)² + (y - 3)² + (0 - 2)²
\(\therefore \)(x - 0)² + (y - 3)² + (0 - 2)² = (x - 0)² + (y - 3)² + (0 - 1)²
=> x²+ y² + 9 - 6y + 4
=x²+y²+1
=> 6y = 12 => y = 2
Putting value of y in (i), we have
2x-3x2=0 => x=3
Thus co-ordinates ofrequired point are (3, 2, 0).